Day 1: Historian Hysteria
Megathread guidelines
- Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
- You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://blocks.programming.dev if you prefer sending it through a URL
FAQ
- What is this?: Here is a post with a large amount of details: https://programming.dev/post/22323136
- Where do I participate?: https://adventofcode.com/
- Is there a leaderboard for the community?: We have a programming.dev leaderboard with the info on how to join in this post: https://programming.dev/post/6631465
Raku
I’m trying warm up to Raku again.
Solution
use v6; sub MAIN($input) { my $file = open $input; grammar LocationList { token TOP { <row>+%"\n" "\n"* } token row { <left=.id> " "+ <right=.id> } token id { \d+ } } my $locations = LocationList.parse($file.slurp); my @rows = $locations<row>.map({ (.<left>.Int, .<right>.Int)}); my $part-one-solution = (@rows[*;0].sort Z- @rows[*;1].sort)».abs.sum; say "part 1: $part-one-solution"; my $rbag = bag(@rows[*;1].sort); my $part-two-solution = @rows[*;0].map({ $_ * $rbag{$_}}).sum; say "part 2: $part-two-solution"; }
I’m happy to see that Lemmy no longer eats Raku code.
Viml
I think viml is a very fun language, i like weird languages lol, so this year im doing it in viml while trying to use as many of the original ed/ex commands as i can (:d, :p, :a, :g, …)
Part 1
!cp ./puzzle1 ./puzzle1.editing e ./puzzle1.editing 1,$sort let row1 = [] g/^\d/let row1 = add(row1, str2nr(expand("<cword>"))) | norm 0dw 1d 1,$sort g/^\d/execute 'norm cc' .. string(abs(expand("<cword>") - row1[line('.') - 1])) $a|---ANSWER--- 0 . 1,$-1g/^\d/call setline("$", str2nr(getline("$")) + str2nr(expand("<cword>")))
Part 2
read ./puzzle1 let cnt = 0 g/^\d/let cnt += expand("<cword>") * \ searchcount(#{pattern: '\s\+' .. expand("<cword>")}).total echo cnt .. "\n" w! ./puzzle1.editing
C#
public class Day01 : Solver { private ImmutableArray<int> left; private ImmutableArray<int> right; public void Presolve(string input) { var pairs = input.Trim().Split("\n").Select(line => Regex.Split(line, @"\s+")); left = pairs.Select(item => int.Parse(item[0])).ToImmutableArray(); right = pairs.Select(item => int.Parse(item[1])).ToImmutableArray(); } public string SolveFirst() => left.Sort().Zip(right.Sort()).Select((pair) => int.Abs(pair.First - pair.Second)).Sum().ToString(); public string SolveSecond() => left.Select((number) => number * right.Where(v => v == number).Count()).Sum().ToString(); }
This is my third program in ruby after the ruby tutorial and half of the rails tutorial, so don’t expect anything too good from it.
Also i did this today since i had time, i will probably not comment every day.
fyi for lines 14-22 you an use
.abs
instead of checking for negatives and.sum
instead of doing it manually. Check my crystal solution to see what I mean
Factor
: get-input ( -- left-list right-list ) "vocab:aoc-2024/01/input.txt" utf8 file-lines [ split-words harvest ] map unzip [ [ string>number ] map ] bi@ ; : part1 ( -- n ) get-input [ sort ] bi@ [ - abs ] 2map-sum ; : part2 ( -- n ) get-input histogram '[ dup _ at 0 or * ] map-sum ;
JavaScript
After writing a procedural to-the-point version in C, tried a JavaScript solution too because it’s just perfect for list comprehension. The part 2 search is inefficient but the data size is small.
Code
const fs = require("fs"); const U = require("./util"); const pairs = fs .readFileSync(process.argv[2] || process.stdin.fd, "utf8") .split("\n") .filter(x => x != "") .map(x => x.split(/ +/).map(Number)); const ls = pairs.map(x => x[0]); ls.sort(); const rs = pairs.map(x => x[1]); rs.sort(); const p1 = U.sum(ls.map((l, i) => Math.abs(l - rs[i]))); const p2 = U.sum(ls.map(l => l * U.count(rs, l))); console.log("01:", p1, p2);
https://github.com/sjmulder/aoc/blob/master/2024/js/day01.js
Part 1 is a sort and a quick loop. Part 2 could be efficient with a lookup table but it was practically instant with a simple non-memoized scan so left it that way.
You are using some interesting techniques there. I never imaged you could use the result of == for adding to a counter.
But how did you handle duplicates in part 2?
I’m not sure if I understand the question correctly but for every number in the left array I count in the right array. That means duplicate work but shrug 😅
Python
Part 1
left_list = [] right_list = [] for line in file: split_line = line.split() left_list.append(int(split_line[0])) right_list.append(int(split_line[1])) sorted_left = sorted(left_list) sorted_right = sorted(right_list) distance = [] for left, right in zip(sorted_left, sorted_right): distance.append(abs(left - right)) total = sum(distance) print(total)
Part 2
file = open('input.txt', 'r') left_list = [] right_list = [] for line in file: split_line = line.split() left_list.append(int(split_line[0])) right_list.append(int(split_line[1])) sim_score = 0 for item in left_list: sim = right_list.count(item) sim_score += (sim * item) print(sim_score)
I am sure there were better ways to do this, this was just the first way in my head, in the order it appeared
Uiua
For entertainment purposes only, I’ll be trying a solution in Uiua each day until it all gets too much for me…
$ 3 4 $ 4 3 $ 2 5 $ 1 3 $ 3 9 $ 3 3 ⊜∘⊸≠@\n # Partition at \n. ⊜(⍆∵⋕)⊸≠@\s # Partition at space, parse ints, sort. &p/+/(⌵-). # Part1 : Get abs differences, sum, print. &p/+×⟜(/+⍉≡⌕)°⊂ # Part 2 : Count instances, mul out, sum, print.
Uiua
Decided to try and use Uiua for each day this year. At least I’m not the only one to get this idea ^^
Run with example input here
PartOne ← ( &rs ∞ &fo "input-1.txt" ⊜(⊜⋕≠@ .)≠@\n. ≡⍆⍉ ⌵/- /+ ) PartTwo ← ( &rs ∞ &fo "input-1.txt" ⊜(⊜⋕≠@ .)≠@\n. ⊢⟜⊣⍉ 0 ⍢(+⊙(:⊙(×⧻⊚◡⌕)↘1⟜⊢)|⋅(≠0⧻)) ⊙(◌◌) # just cleaning up the stack ) &p "Day 1:" &pf "Part 1: " &p PartOne &pf "Part 2: " &p PartTwo
Elixir
defmodule AdventOfCode.Solution.Year2024.Day01 do use AdventOfCode.Solution.SharedParse @impl true def parse(input) do numbers = input |> String.split("\n", trim: true) |> Enum.map(fn l -> String.split(l, ~r/\s+/) |> Enum.map(&String.to_integer/1) end) {Stream.map(numbers, &Enum.at(&1, 0)), Stream.map(numbers, &Enum.at(&1, 1))} end def part1({left, right}) do Enum.zip_reduce(Enum.sort(left), Enum.sort(right), 0, &(&3 + abs(&1 - &2))) end def part2({left, right}) do freq = Enum.frequencies(right) left |> Stream.map(&(&1 * Map.get(freq, &1, 0))) |> Enum.sum() end end
I’ve honestly forgotten how fucking cool the pipe operator is
Rust
I’m doing it in Rust again this year. I stopped keeping up with it after day 3 last year, so let’s hope I last longer this time around.
Solution Spoiler Alert
use std::collections::HashMap; use crate::utils::read_lines; pub fn solution1() { let (mut id_list1, mut id_list2) = get_id_lists(); id_list1.sort(); id_list2.sort(); let total_distance = id_list1 .into_iter() .zip(id_list2) .map(|(left, right)| (left - right).abs()) .sum::<i32>(); println!("Total distance = {total_distance}"); } pub fn solution2() { let (id_list1, id_list2) = get_id_lists(); let id_count_map = id_list2 .into_iter() .fold(HashMap::<_, i32>::new(), |mut map, id| { *map.entry(id).or_default() += 1i32; map }); let similarity_score = id_list1 .into_iter() .map(|id| id * id_count_map.get(&id).copied().unwrap_or_default()) .sum::<i32>(); println!("Similarity score = {similarity_score}"); } fn get_id_lists() -> (Vec<i32>, Vec<i32>) { read_lines("src/day1/input.txt") .map(|line| { let mut ids = line.split_whitespace().map(|id| { id.parse::<i32>() .expect("Ids from input must be valid integers") }); ( ids.next().expect("First Id on line must be present"), ids.next().expect("Second Id on line must be present"), ) }) .unzip() }
python
I didn’t realize it was december until this afternoon. I’ve generally chosen a new or spartan lang to challenge myself, but I’m going easy mode this year with python and just focusing on meeting the requirement.
solution
import aoc def setup(): lines = aoc.get_lines(1) l = [int(x.split()[0]) for x in lines] r = [int(x.split()[1]) for x in lines] return (l, r, 0) def one(): l, r, acc = setup() for p in zip(sorted(l), sorted(r)): acc += abs(p[0] - p[1]) print(acc) def two(): l, r, acc = setup() for n in l: acc += n * r.count(n) print(acc) one() two()